Wednesday, April 16, 2014

We're still working on figuring out a way to geometrically, or verbally, explain the fact that an ellipse is formed by by tracing through the apexes of a family of parabolas that describe the trajectories of a projectile launched with the same initial velocity but different launch angles. Try saying that three times fast, and in the meantime, see picture 1, and refer to the excellent open access article that derives the ellipse mathematically[1].

After I wrote about the elliptical envelope a few days ago +rocktoasted asked if there was a way to describe the construction of the ellipse without using equations or algebra. Hence, the search for a geometrical explanation was born. +Bruce Elliott joined in on the work, and now we have a new collaborator, the only already PhD'ed physicist in the house, my wife Elaine. If you'd like to get in on the collaboration by contributing observations that lead to explaining how to construct the ellipse in [1], please do! Maybe in the end we can right the whole thing up and submit it to one of the +Mathematical Association of America's journals. It might just be the first ad-hoc internet math collaboration to publish there. Who knows??? But, I digress :)

We played around with the diagram of how to make a parabola from the intersection of a plane and set of cones, and came up with the following observations.

Elaine intuited that the base circle shown in picture 2 was actually traced out by the line representing the ground for each parabolic trajectory.

Using the equations from the paper [1], we saw two mathy things immediately. First, if you put 45 degrees, (the angle that gives the maximum range), into the x equation, and 90 degrees, (the angle that gives the maximum height) into the y equation, you'll find that the maximum range is always twice the maximum possible height of the projectile at a given fixed velocity. Bruce saw this first and mentioned it in respect to the eccentricity of the ellipse in the paper always being equal to one half.

The second thing we found requires a bit of calculus. The x range is equal to half of the derivative with respect to theta of the apex height.

I'm not entirely sure what the significance of this yet, but it feels kind of nice from a circular perspective. The x coordinate of a point on a circle is equal to the derivative of the y coordinate of a circle, (picture 4). Notice that the patch factor for the ellipse is just one over the eccentricity, (the ratio of the maximum y distance vs. the maximum x distance of the ellipse).

OK, so so far, we have a family of parabolas formed by laying a plane parallel and tangent to a cone and then pushing that plane through the cone and assuming that the ground is perpendicular to the line that drops straight down from the apex of the cone. Here are the geometry-ish statements that go with this system.

1. When the plane is just tangent to the cone, the parabola that's formed is actually a straight line, (line CK in the diagrams) and represents the trajectory of the projectile if you launched it straight up at a launch angle of 90 degrees.

2. The height of the line CK is proportional to the kinetic energy of the particle at launch, or the potential energy of the particle at the apex, (they're the same for the 90 degree launch angle represented by line CK).

3. As the launch angle sweeps from 90 degrees at CK, to 0 degrees, (the parabola has dwindled to a single point lying at B), the angle around the ground circle centered at M sweeps from 180 degrees at a launch angle of 90 degrees, (line MC) to 0 degrees at a launch angle of 0 degrees, (line MB). When the ground line of the parabola, (line ED) lies along to the diameter of the ground circle M, the projectile has the maximum range. This corresponds to a launch angle of 45 degrees and an angle swept through the ground circle of 90 degrees, (see picture 5).

This feels nice and intuitive, and the three mentioned points certainly match up, but if you'd like a proof of the relations between the launch angle and the angle swept out on the ground circle, see picture 6.

There's one last thing about the distance along the diameter of the ground circle, (line BC), how it's related to the angle theta in the picture above, and how that's related to the maximum height of the projectile in each parabola, but I'll save that for later.